∫ dx 2ax - x^2 = a^n ^ -1 ((x)/(a) -1) . Determine the value of n ....

Let the dimension of $x$ be $[L]$.

The dimension of $dx$ is $[L]$.

For the term $2ax - x^2$ to be dimensionally consistent, the dimension of $a$ must be $[L]$.

The dimension of the denominator $\sqrt{2ax - x^2}$ is $\sqrt{[L^2]} = [L]$.

Therefore, the dimension of the left-hand side is $\dfrac{[L]}{[L]} = [L^0]$.

On the right-hand side, the argument of the inverse sine function, $\dfrac{x}{a} - 1$, is dimensionless. The function $\sin^{-1}$ yields an angle, which is also dimensionless.

Thus, the dimension of the right-hand side is $[a^n] = [L]^n$.

Equating the dimensions of both sides, we get $[L^0] = [L]^n$, which implies $n = 0$.

Alternatively, evaluating the integral directly:

$\int \dfrac{dx}{\sqrt{2ax - x^2}} = \int \dfrac{dx}{\sqrt{a^2 - (x^2 - 2ax + a^2)}} = \int \dfrac{dx}{\sqrt{a^2 - (x - a)^2}}$

Using the standard formula $\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}$\left(\dfrac{x}{a}\right), we get:

\int \dfrac{dx}{\sqrt{a^2 - (x - a)^2}} = \sin^{-1} \left(\dfrac{x - a}{a}\right) = \sin^{-1}\left(\dfrac{x}{a} - 1\right)$<br><br>Comparing this with$a^n \sin^{-1} \left(\dfrac{x}{a} -1\right)$, we find$a^n = 1$, which gives$n = 0$.$